Killing vecotrs of Schwarzschild metric

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Discussion Overview

The discussion focuses on the search for Killing vectors in the context of the Schwarzschild metric, exploring the possibility of transforming the metric to identify additional Killing vectors beyond the known ones. Participants examine the implications of spherical symmetry and the nature of transformations applied to the metric.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that since the Schwarzschild metric is independent of certain coordinates, it may be possible to transform the metric to find additional Killing vectors in the spherical part of the metric.
  • Another participant asserts that such a transformation is not possible and points out that the generators of rotations, which form the basis of ##so(3)##, are also Killing fields due to the spherical symmetry of the Schwarzschild metric.
  • A participant questions whether knowing all transformations is necessary, suggesting that a differential form could suffice to derive Killing vectors.
  • Another participant argues that if the metric could be transformed to the form ##d\alpha^2 + d\beta^2##, it would imply a flat geometry, which contradicts the spherical nature of the Schwarzschild metric.
  • One participant demonstrates that their proposed transformation does not yield the correct metric for the 2-sphere, reinforcing the claim that the desired transformation is not feasible.

Areas of Agreement / Disagreement

Participants generally disagree on the possibility of transforming the Schwarzschild metric to identify additional Killing vectors. There is no consensus on the validity of the proposed transformation methods.

Contextual Notes

Participants express uncertainty regarding the assumptions underlying the transformations and the implications of spherical symmetry on the existence of Killing vectors.

Vrbic
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Hello,
I have a question, whether is possible to looking for Killing Vectors (KV) in this way (I know about general solution):[itex][/itex]
From Schwarzschild metric I can see two KV [itex]\frac{\partial}{\partial t}[/itex] and [itex]\frac{\partial}{\partial\phi}[/itex]. Then I see that other trivial KV arent there. Metric in dt and dr is independent on [itex]\theta, \phi[/itex] so I suppose I can "split" metric and looking for KV just in spherical part [itex]d\theta^2+\sin^2{\theta}d\phi^2[/itex].
Can I suppose transformation this metric to the form: [itex]d\alpha^2+d\beta^2[/itex] and claim the [itex]\frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta}[/itex] are KV?
 
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No. You won't be able to find such a transformation.

Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of ##so(3)##, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.
 
WannabeNewton said:
No. You won't be able to find such a transformation.

Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of ##so(3)##, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.

Do I need know all transformation? I thought that differencial form of it could be enough. If I suppose [itex]\alpha=\alpha(\theta,\phi), \beta=\beta(\theta,\phi)[/itex], than I can write [itex]d\alpha=\alpha_{\theta}+\alpha_{\phi}[/itex], [itex]d\beta=\beta_{\theta}+\beta_{\phi}[/itex], where subscript means differentation. When I put it back to the [itex]ds^2=d\alpha^2+d\beta^2[/itex] I get 3 eq.:
[itex]1=\alpha_{\theta}^2+\beta_{\theta}^2[/itex]
[itex]0=\alpha_{\theta}\alpha_{\phi}+\beta_{\theta}\beta_{\phi}[/itex]
[itex]\sin^2{\theta}=\alpha_{\phi}^2+\beta_{\phi}^2[/itex].
From them (if from first guess [itex]\alpha_{\theta}=\cos{\phi}[/itex], [itex]\beta_{\theta}=\sin{\phi}[/itex]) I get [itex]d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi[/itex], [itex]d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi[/itex].
Multiplying by [itex]g^{\mu\nu}[/itex] I hope can get [itex]\frac{\partial}{\partial\alpha}[/itex], [itex]\frac{\partial}{\partial\beta}[/itex]. What is wrong?
 
Vrbic said:
Can I suppose transformation this metric to the form: [itex]d\alpha^2+d\beta^2[/itex] and claim the [itex]\frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta}[/itex] are KV?
If you could put the metric in that form, it would be FLAT. It's not flat, it's a sphere.
 
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Vrbic said:
From them (if from first guess [itex]\alpha_{\theta}=\cos{\phi}[/itex], [itex]\beta_{\theta}=\sin{\phi}[/itex]) I get [itex]d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi[/itex], [itex]d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi[/itex].

This clearly doesn't reduce to the metric on the 2-sphere: simple substitution yields ##d\alpha^2 + d\beta^2 = d\theta ^2 + \sin^2 \theta d\phi^2 + (\sin \theta \cos\phi - \sin 2\phi )\sin\theta d\theta d\phi##.

You simply won't be able to put the 2-sphere metric in the form ##ds^2 = d\alpha^2 + d\beta^2## and Bill has elucidated above why this is impossible.
 
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